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Signal questions......

  • 09-01-2011 3:45am
    #1
    Registered Users, Registered Users 2 Posts: 1,771 ✭✭✭


    Im kinda curious,if someone was 3048 Meters up on a mountain HOW FAR AWAY COULD THEY TALK TO SOMEONE WHO IS ALSO 3048 Meters on a mountain? (NO ACTIVE SKIP)

    Say someone on the east was 3048 Meters on a mountain and someone far far away was 3048 Meters up,could they here each other seeing they are at the same elevation?

    Now the lower the freq they are on the better chance obviously but regardless,at 3048 Meters I RECKON ANY FREQ CAN BE HEARD WAY FAR AWAY (@ the same elevation)

    Just something i was curious of.......


Comments

  • Registered Users, Registered Users 2 Posts: 32,417 ✭✭✭✭watty


    At say 50MHz the distance is MUCH more than 1200MHz.

    If there is a mountain half way of correct height you can get you can get much more distance.

    As the height increases you reach a point where the distance will hardly increase at all.

    Also there is still inverse square law. double distance gives 1/4 power. So depends on power.


  • Registered Users, Registered Users 2 Posts: 177 ✭✭brownmini


    Dude111 wrote: »
    Im kinda curious,if someone was 3048 Meters
    up on a mountain HOW FAR AWAY COULD THEY TALK TO SOMEONE WHO IS
    ALSO 3048 Meters on a mountain? (NO ACTIVE SKIP)

    Say someone on the east was 3048 Meters on a mountain and someone far
    far away was 3048 Meters up,could they here each other seeing they are
    at the same elevation?

    Now the lower the freq they are on the better chance obviously but
    regardless,at 3048 Meters I RECKON ANY FREQ CAN BE HEARD WAY FAR
    AWAY (@ the same elevation)

    The old formula for calculating distance (in miles) line of sight used to be
    1.23 x SQUARE_ROOT( height_of_TX_in_feet + height_of_RX_in_feet)

    Which gives a distance of 173 miles


    However if you have a gawk at
    http://www.qsl.net/kd4sai/distance.html

    And type in your values in feet, this website seems to suggest that the
    total range of two boyos sitting at 9999 feet is 282 miles with each of
    them capable of reaching 141 miles by themselves.

    There is more on Wkipedia at http://en.wikipedia.org/wiki/Radio_horizon

    Remember this is theory, may not work in practice.
    If you are going to give it a try and if you are going to try it in Ireland then
    good luck in finding even one place that is over 5000 feet high never mind
    2 at 9999 feet high and especially with no obstacles in the way at the mid
    point.


  • Registered Users, Registered Users 2 Posts: 32,417 ✭✭✭✭watty


    Any decent formula should have a frequency component. Most "radio horizon" calculators are copy of WWII radar calculation with empirical constant for a high UHF frequency, so too big a number for 10GHz and too small for 30MHz. Also doesn't take into account Fresnel effects or Ground wave on LF, MF, HF and low VHF

    A mid point obstacle can increase range! (Diffraction).

    always do "view source"
    via http://www.qsl.net/w4sat/horizon.htm
    <title>VHF/UHF Line of Sight Calculator</title>
    <script language="JavaScript">
    <!--
    function calc(page) {
       mast1 = parseFloat(page.height1.value);
       mast2 = parseFloat(page.height2.value);
       if(page.units[1].checked) {
          mast1 = mast1/0.3048;
          mast2 = mast2/0.3048;
       }
       d1 = Math.sqrt(2*mast1);
       d2 = Math.sqrt(2*mast2);
       if(page.dist[1].checked) {
          d1 = parseFloat(d1*1.609);
          d2 = parseFloat(d2*1.609);
          unit = "   Km";
       }  else unit = "   Miles";
       d1 = Math.round(d1);
       d2 = Math.round(d2);
       page.horiz1.value = d1+unit;
       page.horiz2.value = d2+unit
       page.distance.value = Math.round(d1 + d2)+unit;
    }
    //-->
    </script>
    </head>
    

    http://www.qsl.net/kd4sai/distance.html
    <script language="JavaScript">
    <!--
    function calc(page) {
       mast1 = parseFloat(page.height1.value);
       mast2 = parseFloat(page.height2.value);
       if(page.units[1].checked) {
          mast1 = mast1/0.3048;
          mast2 = mast2/0.3048;
       }
       d1 = Math.sqrt(2*mast1);
       d2 = Math.sqrt(2*mast2);
       if(page.dist[1].checked) {
          d1 = parseFloat(d1*1.609);
          d2 = parseFloat(d2*1.609);
          unit = "   Km";
       }  else unit = "   Miles";
       d1 = Math.round(d1);
       d2 = Math.round(d2);
       page.horiz1.value = d1+unit;
       page.horiz2.value = d2+unit
       page.distance.value = Math.round(d1 + d2)+unit;
    }
    //-->
    </script>
    

    Both those are not much good. They both copied something older.
    This is the same and equally poor http://radarproblems.com/calculators/horizon.htm

    See here. Some allowance for Frequency. http://www.fcc.gov/mb/audio/bickel/curves.html
    At lower HF frequencies you also get Groundwave. This is NOT the same as LOS communications. Even 28Mhz has some Groundwave.
    198kHz has lots ;)

    See http://www.smeter.net/propagation/grndwav3.php For 20MHz
    Distance to the radio horizon, approximately 80/Cube-root(MHz) Km, is where diffraction loss begins to take effect. Up to the horizon the Earth is assumed flat. Diffraction loss in dB increases in proportion to path length and to the cube-root of frequency. It is also affected by the type of terrain.

    download Groundwave calculator (takes frequency into account) http://www.smeter.net/software/grndwav3.exe


  • Registered Users, Registered Users 2 Posts: 177 ✭✭brownmini


    watty wrote: »
    Any decent formula should have a frequency component. Most "radio horizon" calculators are copy of WWII radar ...

    One must learn to walk before one considers to run.


  • Registered Users, Registered Users 2 Posts: 1,771 ✭✭✭Dude111


    watty wrote:
    198kHz has lots ;)
    Yes i reckon it does!! (Anything below .500Mhz has excellent coverage (With less power)


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